`x ^(2) +y ^(2) =a ^(2)` is the standard equation of a circle centred at `(0,0)` and radius is a.
Paramatric Equation to Standard Circle:
Parametric Equation for the circle `x^(2) +y ^(2) =a ^(2) ` is `x =a cos theta, y =a sin theta.`
Director Circle:
Director circle is the locus of point of intersection of two perpendicular tangents.
Two points `A (-30^(@)) and B (150^(@))` lies on circle `x ^(2) +y ^(2) =9.` The point `(theta)` which moves on a circle such that area of `Delta PAB` is maximum, is/are

Options

A. `60^(@)+2kpi, k in l`
B. `120^(@)+2kpi, k in l`
C. `75^(@)+2kpi , k in l`
D. none of these

Answer

To solve the problem, we need to find the point \( P \) on the circle \( x^2 + y^2 = 9 \) such that the area of triangle \( PAB \) is maximized, where points \( A \) and \( B \) are given as \( A(-3\sqrt{3}, -3) \) and \( B(3\sqrt{3}, 3) \).

### Step-by-Step Solution:

1. **Identify the Circle and Points**:
The equation of the circle is \( x^2 + y^2 = 9 \), which has a center at \( (0, 0) \) and a radius \( r = 3 \). The points \( A \) and \( B \) are located on this circle.

2. **Parametric Equations**:
The parametric equations for the circle can be expressed as:
\[ x = 3 \cos \theta, \quad y = 3 \sin \theta \]
where \( \theta \) is the angle parameter.

3. **Area of Triangle \( PAB \)**:
The area \( A \) of triangle \( PAB \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting \( P(3 \cos \theta, 3 \sin \theta) \), \( A(-3\sqrt{3}, -3) \), and \( B(3\sqrt{3}, 3) \):
\[ \text{Area} = \frac{1}{2} \left| 3 \cos \theta \left(-3 - 3\right) + (-3\sqrt{3})(3 - 3 \sin \theta) + (3\sqrt{3})(3 \sin \theta + 3) \right| \]

4. **Simplifying the Area Expression**:
Simplifying the expression:
\[ = \frac{1}{2} \left| 3 \cos \theta (-6) + (-3\sqrt{3})(3 - 3 \sin \theta) + 3\sqrt{3}(3 \sin \theta + 3) \right| \]
\[ = \frac{1}{2} \left| -18 \cos \theta + 9\sqrt{3} \sin \theta + 9\sqrt{3} \right| \]

5. **Finding Maximum Area**:
To maximize the area, we can differentiate the area expression with respect to \( \theta \) and set the derivative to zero. However, a more straightforward approach is to recognize that the area is maximized when the height from point \( P \) to line \( AB \) is maximized.

6. **Using Geometry**:
The maximum area occurs when \( P \) is perpendicular to the line segment \( AB \). The slope of line \( AB \) can be found, and the angle \( \theta \) can be determined such that \( P \) is directly above or below the midpoint of \( AB \).

7. **Calculating the Midpoint**:
The midpoint \( M \) of \( AB \) is:
\[ M = \left( \frac{-3\sqrt{3} + 3\sqrt{3}}{2}, \frac{-3 + 3}{2} \right) = (0, 0) \]
Since \( P \) must lie on the circle, we can find the angles corresponding to the maximum area.

8. **Finding Angles**:
The angles that maximize the area can be determined using the tangent function:
\[ \tan \theta = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}} \implies \theta = 60^\circ \text{ or } 240^\circ \]

### Final Answer:
The angles \( \theta \) that maximize the area of triangle \( PAB \) are:
\[ \theta = 60^\circ \text{ and } 240^\circ \]